Chapter 05

CONTENTS

1 IDEAL GASES

1.1 Boyle's Law

1.2 Charles' Law

1.3 Avogadro's Law

1.4 The Equation of State For an Ideal Gas

1.5 The Density of an Ideal Gas

1.6 Standard Conditions

1.7 Mixtures of Ideal Gases

1.7.1 Dalton's Law of Partial Pressures

1.7.2 Amagat's Law

1.8 Apparent Molecular Weight

1.9 Specific Gravity of a Gas

2 BEHAVIOUR OF REAL GASES

2.1 Compressibility Factor For Natural Gases

2.2 Law of Corresponding States

2.3 Pseudocritical Properties of Natural Gases

2.4 Impact of Nonhydrocarbon Components on

z Value

2.5 Standard Conditions For Real Reservoir

Gases

3 GAS FORMATION VOLUME FACTOR

4 COEFFICIENT OF ISOTHERMAL

COMPRESSIBILITY OF GASES

5 VISCOSITY OF GASES

5.1 Viscosity

5.2 Viscosity of Mixtures

6 EQUATIONS OF STATE

6.1 Other Equations-of-State

6.2 Van de Waals Equation

6.3 Benedict - Webb - Rubin Equation (BWR)

6.4 Redlich - Kwong Equation

6.5 Soave, Redlich Kwong Equation

6.6 Peng Robinson Equation of State

6.7 Application to Mixtures

Behaviour of Gases

2

LEARNING OBJECTIVES

Having worked through this chapter the Student will be able to:

• Present the ideal equation of state, PV=nRT.

• Calculate the mass of an ideal gas given PV 7T values.

• Derive an equation to calculate the density of an ideal gas.

• Convert a mixture composition between weight and mole fraction.

• Present an equation and calculate the apparent molecular weight of a mixture.

• Define and calculate the specific gravity of a gas.

• Present the equation of state, EOS, for a ʻreal gas ̓and explain what ʻZ ̓ is,

PV=ZnRT.

• Define the pseudocritical pressure and psuedocritical temperature and be able

to use them to determine the ʻZ ̓value for a gas mixture.

• Express and calculate reservoir gas volumes in terms of standard cubic

volumes.

• Define the gas formation volume factor and derive an equation fore it using the

EOS.

• Calculate the volume of gas in a reservoir in terms of standard cubic volumes

given prerequisite data.

• Calculate the viscosity of a gas of a specific composition given perquisite

equations and figures.

• Be aware of the development of EOSʼs to predict reservoir fluid properties.

Institute of Petroleum Engineering, Heriot-Watt University 3

INTRODUCTION

A gas is a homogenous fluid that has no definite volume but fills completely the vessel

in which it is placed. The system behaviour of gases is vital to petroleum engineers

and the laws governing their behaviour should be understood. For simple gases these

laws are straightforward but the behaviour of actual hydrocarbon gases particularly

at the conditions occurring in the reservoir are more complicated.

We will review the laws that relate to the pressure, volume and temperatures of gases

and the associated equations. These relationships were previously termed gas laws;

it is now more common to describe them as equations of state.

1 IDEAL GASES

The laws relating to gases are straightforward in that the relationships of pressure,

temperature and pressure are covered by one equation. First consider an ideal gas.

An ideal gas is one where the following assumptions hold:

• Volume of the molecules i.e. insignificant with respect to the total volume of

the gas.

• There are no attractive or repulsive forces between molecules or between

molecules and container walls.

• There is no internal energy loss when molecules collide.

Out of these assumptions come the following equations.

1.1 Boyleʼs LawAt constant temperature the pressure of a given weight of a gas is inversely proportional

to the volume of a gas.

i.e.

V

1

P or PV = constant, T is constant

(1)

P = pressure, V = volume, T = temperature.

1.2 Charles ̓LawAt constant pressure, the volume of a given weight of gas varies directly with the

temperature:

i.e.

V T or

V

T = constant, P is constant

(2)

The pressure and temperature in both laws are in absolute units.

Behaviour of Gases

4

1.3 Avogadroʼs LawAvogadroʼs Law can be stated as: under the same conditions of temperature and

pressure equal volumes of all ideal gases contain the same number of molecules. That

is, one molecular weight of any ideal gas occupies the same volume as the molecular

weight of another ideal gas at a given temperature and pressure.

Specifically, these are:

(i) 2.73 x 1026 molecules/lb mole of ideal gas.

(ii) One molecular weight (in lbs) of any ideal gas at 60˚F and 14.7 psia

occupies a volume of 379.4 cu ft.

One mole of a material is a quantity of that material whose mass in the unit system

selected is numerically equal to the molecular weight.

eg. one lb mole of methane CH4 = 16 lb

one kg mole of methane CH4 = 16kg

1.4 The Equation of State for an Ideal GasBy combining the above laws an equation of state relating pressure, temperature and

volume of a gas is obtained.

PV

T constant

(3)

R is the constant when the quantity of gas is equal to one mole.

It is termed the Universal Gas Constant and has different values depending on the

unit system used, so that;

R in oilfield units = 10 732.

cu ft psia

lb mole R

Table 1 gives the values for different unit systems.

P V T n R � �

�

psia �� cu ft �� R �� lb - mole � 10.73 ��

atm �� cu ft �� K �� lb - mole � 1.3145 �

atm �� cc �� K �� gm - mole � 82.06 �

atm �� litre �� K �� gm - mole � 0.08206 �

atm �� cu ft �� R �� lb - mole � 0.730 �

mm Hg � litre �� K �� gm - mole � 62.37 �

in.Hg �� cu ft �� R �� lb - mole � 21.85 �

�

Table 1 Values of R for different unit systems


For n moles the equation becomes:

PV = nRT (4)

T= absolute temperature oK or oR where

ºK=273 +oC and oR=460 +oF

To find the volume occupied by a quantity of gas when the conditions of temperature

and pressure are changed from state 1 to state 2 we note that:

n PV

RT is a constant so that

P V

T =

P V

T

1 1

1

2 2

2

EXERCISE 1.

A gas cylinder contains methane at 1000 psia and 70°F. If the cylinder has a vol-ume of 3 cu.ft assuming methane is an ideal gas calculate the mass of methane in

the cylinder.

1.5 The Density of an Ideal GasSince density is defined as the weight per unit volume, the ideal gas law can be used

to calculate densities.

g = weight / volume =

m

V

where g is the gas density

For 1 mole m = MW MW = Molecular weight

V RT

P

= MW.P

RTg

(5)

EXERCISE 2.

Calculate the density of the gas in the cylinder in exercise 1.

Behaviour of Gases

6

1.6 Standard ConditionsOil and gas at reservoir conditions clearly occur under a whole range of temperatures

and pressures.

It is common practice to relate volumes to conditions at surface, ie 14.7 psia and

60˚F.

ie

P V

T

P V

T

res res

res

sc sc

sc (6)

sc - standard conditions res - reservoir conditions

This relationship assumes that reservoir properties behave as ideal. This is NOT the

case as will be discussed later.

EXERCISE 3.

Assuming methane is at the conditions of exercise 1, calculate the volume the gas would occupy at standard conditions.

1.7 Mixtures of Ideal GasesPetroleum engineering is concerned not with single component gases but mixtures

of a number of gases.

Laws established over early years governing ideal gas mixtures include Daltonʼs

Law and Amagatʼs Law.

1.7.1 Daltonʼs Law of Partial PressuresThe total pressure exerted by a mixture of gases is equal to the sum of the pressures

exerted by its components. The partial pressure is the contribution to pressure of

the individual component.

Consider a gas made up of components A, B, C etc

The total pressure of the system is the sum of the partial pressures

ie

P = P + P + P + .....A B C (7)

where A, B and C are components.

therefore


P = n RT

V n

RT

V n

RT

V

i.e. P = RT

Vn

P

P =

n

n = y

A B C

j

j j

j

(8)

where yj = mole fraction of jth component.

The pressure contribution of a component, its partial pressure, is the total pressure

times the mole fraction.

1.7.2 Amagatʼs LawAmagatʼs Law states that the volume occupied by an ideal gas mixture is equal to the

sum of the volumes that the pure components would occupy at the same temperature

and pressure. Sometimes called the law of additive volumes.

i.e.

V = V + V + VA B C (9)

V = n RT

P + n

RT

P + n

RT

P

V = RT

Pn

V

V =

n

n = y

A B C

j

j j

ji e. . (10)

i.e, for an ideal gas the volume fraction is equal to the mole fraction.

It is conventional to describe the compositions of hydrocarbon fluids in mole terms.

This is because of the above laws. In some circumstances however weight compositions

might be used as the basis and it is straight forward to convert between the two.

EXERCISE 4.

A gas is made up of the following components; 25lb of methane, 3 lb of ethane and 1.5 lb of propane. Express the composition of the gas in weight and mole fractions.

Behaviour of Gases

8

1.8 Apparent Molecular WeightA mixture does not have a molecular weight although it behaves as though it had a

molecular weight. This is called the apparent molecular weight. AMW

If yj represents the mole fraction of the jth component:

AMW = y MWj j

AMW for air = 28.97, a value of 29.0 is usually sufficiently accurate.

EXERCISE 5.

What is the apparent molecular weight of the gas in exercise 4

1.9 Specific Gravity of a GasThe specific gravity of a gas,

g is the ratio of the density of the gas relative to that of

dry air at the same conditions.

g

g

air

=

(11)

Assuming that the gases and air are ideal.

g

g

air

g

air

g =

M P

RTM P

RT

= M

M =

M

29

Mg = AMW of mixture, Mair = AMW of air.

EXERCISE 6.

What is the gas gravity of the gas in exercise 4 ?

2 BEHAVIOUR OF REAL GASES

The equations so far listed apply basically to ideal systems. In reality, however,

particularly at high pressures and low temperatures the volume of the molecules are

no longer negligible and attractive forces on the molecules are significant.


The ideal gas law, therefore, is not too applicable to light hydrocarbons and their

associated fluids and it is necessary to use a more refined equation.

There are two general methods of correcting the ideal gas law equation:

(1) By using a correction factor in the equation PV = nRT

(2) By using another equation-of-state

2.1 Compressibility Factor for Natural GasesThe correction factor ʻz ̓which is a function of the gas composition, pressure and

temperature is used to modify the ideal gas law to:

PV = znRT (12)

where the factor ̒ z ̓is known as the compressibility factor and the equation is known

as the compressibility equation-of-state or the compressibility equation.

The compressibility factor is not a constant but varies with changes in gas composition,

temperature and pressure and must be determined experimentally (Figure 1).

To compare two states the law now takes the form:

P V

z T =

P V

z T

1 1

1 1

2 2

2 2 (13)

z is an expression of the actual volume to what the ideal volume would be.

i.e.

z

Vactual

Videal

(14)

Tem

pera

ture

= c

onst

ant

00

0.5

1.0

PRESSURE, P

Co

mp

ressib

ility

fa

cto

r, Z

Figure 1 Typical plot of the compressibility factor as a function of pressure at constant

temperature.

Behaviour of Gases

10

Although all gases have similar shapes with respect to z the actual values are component

specific. However through the law of corresponding states all pure gases are shown

to have common values.

2.2 Law of Corresponding StatesThe law of corresponding states shows that the properties of many pure liquids and

gases have the same value at the same reduced temperature (Tr) and pressure (P

r)

where:

T = T

T and P =

P

Pr

c

r

c (15)

Where, Tc and P

c are the pure component critical temperature and pressure.

The compressibility factor ʻz ̓follows this law. It is usually presented vs Tr and Pr.

Although in many cases pure gases follow the Law of Corresponding States, the gases

associated with hydrocarbon reservoirs do not. The Law has however been used to

apply to mixtures by defining parameters called pseudo critical temperature and

pseudocritical pressure .

For mixtures a pseudocritical temperature and pressure, Tpc

and Ppc

is used such

that:

T = y T and P = y Ppc j cj pc j cj (16)

where y is the mole fraction of component j and Tcj and P

cj are the critical temperature

and pressure of component j.

It should be emphasised that these pseudo critical temperature and pseudocritical

pressures are not the same as the real critical temperature and pressure. By definition

the pseudo values must lie between the extreme critical values of the pure components

whereas the actual critical values for mixtures can be outside these limits, as was

observed in the Phase Behaviour chapter.

EXERCISE 7.

Calculate the pseudo critical temperature and pseudocritical pressure of the mixture in exercise 4 .

For mixtures the compressibility factor (z) has been generated with respect to natural

gases 1, where ʻz ̓ is plotted as a function of pseudo reduced temperature, Tpr and

pseudo reduced pressure Ppr where


0

1.0

1.1

0.9

0.8

0.7

0.6

0.5

0.4

0.3

0.25

1.1

1.0

0.9

1.01.05

1.05

1.1

1.2

1.3

1.4

1.5

1.6 1.7

1.8 1.9

2.0 2.2

2.4

2.6

3.0

3.02.8

1.21.3

1.1

1.10.95

1.7

1.6

1.5

1.4

1.3

1.2

1.1

1.0

0.9

1 2 3 4 5 6 7 8

7 8 9 10 11 12 13 14 15

Compressibility of

Natural Gases

(Jan. 1, 1941)

Compressibility Factors for Natural Gases as a

Function of Pseudoreduced Pressure and Temperature.

Co

mp

ressib

ility

Fa

cto

r, z

Pseudo Reduced Temperature

Pseudo Reduced Pressure, Pr

3.02.82.62.42.2

2.01.91.8

1.7

1.6

1.5

1.45

1.35

1.4

1.3

1.25

1.2

1.15

1.1

2.6 2.4

2.22.0 1.9

1.71.6 1.4

1.3

1.2

1.1

1.05

1.05

1.8

1.4

1.5

Pseudo Reduced Pressure, Pr

Figure 2 Compressibility factors for natural gas1 (Standing & Katz, Trans AIME, 1942)

Behaviour of Gases

12

TT

Tand

P

Ppr

pc pc

Ppr

(17)

The use of this chart , figure 2 ,has become common practise to generate z values for

natural gases. Poettmann and Carpenter 2 have also converted the chart to a table.

Various equations have also been generated based on the tables.

EXERCISE 8.

For the gas of exercise 4 determine the compressibility factor at a temperature of 150°F and a pressure of 3500psia.

2.3 Pseudocritical Properties of Natural GasesThe pseudocritical properties of gases can be computed from the basic composition

but can also be estimated from the gas gravity using the correlation presented in

Figure 3.

0.5 0.6 0.7 0.8 0.9 1.0 1.1 1.2

Pseudocritical Properties of Natural Gases

Pseudocritical Tem

pera

ture

, R

Pseudocritical P

ressure

, psia

Gas Gravity (air = 1)

700

650

600

550

500

450

400

350

300

Condensate Well Fluids

Miscellaneous Gases

Miscellaneous G

ases

Condensate Well F

luids

Figure 3 Pseudocritical properties of natural gases 3


2.4 Impact of Nonhydrocarbon Components on z value.Components like hydrogen sulphide, and carbon dioxide have a significant impact

on the value of z. If the method previously applied is used large errors in z result.

Wichert and Aziz 4 have produced an equation which enables the impact of these

two gases to be calculated.

T'pc

= Tpc

- (18)

and

pp T

T y ypc

pc pc

pc H S H S2 21

(19)

T'pc

and p'pc

are used to calculate Tpr and p

pr. The value for is obtained from

the figure 4 from the Wichert and Aziz paper

0 10 20 30 40 50 60 70 800

10

20

30

40

50

60

70

80

PER CENT H2S

PE

R C

EN

T C

02

510

15

20

25

30

15

20

25

30

E

34.5

Figure 4 Adjustment factors for pseudocritiacl properties for non hydrocarbon

gases(Wichert & Aziz)

Behaviour of Gases

14

EXERCISE 9.

Calculate the pseudo critical properties of the gas in exercise 4 if it also contained 3

lb of hydrogen sulphide, 10lb of carbon dioxide and 2.5lb of nitrogen

1

2

3

Gas

Components

Mol

weight

Mole

fraction

Pc-psi Tc °R Ppcpsia

Methane 25 0.56 16.04 0.035 0.743 667.00 344 495.8 255.70

Ethane 3 0.07 30.07 0.002 0.048 708.00 550 33.7 26.17

Propane 1.5 0.03 44.09 0.001 0.016 616.00 666 10.0 10.81

Hydrogen 3 0.07 34.08 0.002 0.042 1306 673 54.8 28.25

sulphide

Carbon 10 0.22 44.01 0.005 0.108 1071 548 116.1 59.38

Dioxide

Nitrigen 2.5 0.06 28.02 0.002 0.043 493 227 21.0 9.66

Total 45 1.00 0.0466 1.000 731 390

TpcWeight Wgt

fraction

lb moles

4

5

6

From Wichert & Azis chart for compositions of H2S and CO

2 = 19

T = T - = 371 R

P = 694.3

pc pc

o

pc

pp T

T y ypc

pc pc

pc H S H S2 21

2.5 Standard Conditions for Real Reservoir GasesAs indicated in section 1.6 for ideal gases it is convenient to describe the quantity of

gas to a common basis and this is termed the standard conditions, giving rise to the

standard cubic foot and the standard cubic metre. The petroleum engineer is primarily

interested in volume calculations for gaseous mixtures. Throughout the industry gas

volumes are measured at a standard temperature of 60˚F (15.6˚C) and at a pressure of

14.7 psia (one atmosphere). These conditions are referred to as standard temperature

and pressure STP. Standard Cubic Feet, the unit of volume measured under these

conditions is sometimes abbreviated SCF or scf (SCM is Standard Cubic Metres). It

is helpful to consider these expressions not as volumes but as an alternate expression

of the quantity of material. For example a mass of gas can be expressed as so many

standard cubic feet or metres.

EXERCISE 10.

Express the quantity of 1 lb mole of a gas as standard cubic feet.


EXERCISE 11.

Express the mass of gas in exercise 4 as standard cubic feet.

3 GAS FORMATION VOLUME FACTOR

The petroleum industry expresses its reservoir quantities at a common basis of surface

conditions which for gases is standard cubic volumes. To convert reservoir volumes

to surface volumes the industry uses formation volume factors. For gases we have

Bg, the gas formation volume factor, which is the ratio of the volume occupied at

reservoir temperature and pressure by a certain weight of gas to the volume occupied

by the same weight of gas at standard conditions. The shape of Bg as a function of

pressure is shown in figure 5.

Bvolume occupied at reservoir temperature and pressure

volume occupied at STPg

The gas formation volume factor can be obtained from PVT measurements on a gas

sample or it may be calculated from the equations-of-state discussed previously.

One definition of the gas formation volume factor is: it is the volume in barrels

that one standard cubic foot of gas will occupy as free gas in the reservoir at the

prevailing reservoir pressure and temperature.

Depending on the definition the units will change and the units will be; rb free

gas/scf gas or rm3 free gas/scm gas

.008

.006

.004

.002

1000 2000 3000

Bg

rb/scf

PRESSURE (psig)

Figure 5 Gas Formation Volume Factor, Bg

Behaviour of Gases

16

For example Bg for a reservoir at condition 2 is;

BV

V

P T z

P T zg

2

sc

sc 2 2

2 sc sc (20)

ʻsc ̓refers to standard conditions. z at standard conditions is taken as 1.0

The reciprocal of Bg is often used to calculate volumes at surface so as to reduce the

possibility of misplacing the decimal point associated with the values of Bg being

less than 0.01, ie:

volume at surface

volume in formation Bg

1E

E is sometimes referred to as the expansion factor.

Usually the units of Bg are barrels of gas at reservoir conditions per standard cubic

foot of gas, ie bbl/SCF or cubic metres per standard cubic metre.

BV

Vg

R

sc (21)

R and sc are reservoir and standard conditions respectively.

V

znRT

PR

(22)

T and P at reservoir conditions:

Vz nRT

Psc

sc sc

sc (23)

z = 1 for standard conditions

B zT

T

P

P

cu. ft

SCFg

sc

sc. .

(24)

Since Tsc = 520˚Rm Psc = 14.7 psia for most cases

B 0

zT

P

cu. ft

SCFg .0283

B 0 zT

P

cu. ft

SCF

bbl

5.615 cu ftg .0283

or


B 0

zT

P

res bbl

SCFg .00504

(25)

EXERCISE 12.

Calculate the gas formation factor for a gas with the composition of exercise 4 existing at the reservoir conditions given in exercise 8.

EXERCISE 13.

A reservoir exists at a temperature of 150°F (as for exercise 8) suitable for storing gas. It has an areal size of 5 miles by 2 miles and is 200ft thick. The average porosity is 20% and there is no water present. How much gas of the composition of exercise 4 can be stored at a pressure the same as in exercise 8 i.e. 3500 psia ? (1 mile= 5280 ft.)

4 COEFFICIENT OF ISOTHERMAL COMPRESSIBILITY OF GAS-

ES

The compressibility factor, z, must not be confused with the compressibility which is

defined as the change in volume per unit volume for a unit change in pressure, or

cV

V

P or

V

V

Pg

m

m1 1

(26)

Vm is the specific volume or volume per mole.

cg is not the same as z, the compressibility factor.

For an ideal gas:

PV = nRT or:

dV

dP

nRT

P

c = 1

V

nRT

P=

1

P

2

g 2

(27)

For real gases:

V =

znRT

P

Behaviour of Gases

18

V

PnRT

Pdz

dP

P

cP

nRTz

nRT

PP

z

Pz

cP

1

z

z

P

T

2

g 2

g

z

1.

(28)

dz/dP can be obtained from the slope of the z vs P curve.

The Law of Corresponding states can be used to express the above equation in

another form

P = P P

z

P

P

P

z

P

z

P

z

P

pc pr

pr

pr

pr

P

P P

P

pr

pc

pc

1

1

Combining this equation with eqn 28 above yields

cP P

1

zP

z

P

c PP

1

z

z

P

g

pc pr pc pr

g pc

pr pr

1

1

T

T

pr

pr (29)

Units of cg = P-1, and c

gP

c is dimensionless

cpP

pc is called pseudo reduced compressibility, c

pr


Since the pseudo reduced compressibility is a function of ʻz ̓and pseudo reduced

pressure, the graph of Figure 2 can be used with Equation 29 to calculate values of

cpr.

5 VISCOSITY OF GASES

5.1 ViscosityViscosity is a measure of the resistance to flow. It is given in units of centipoise.

A centipoise is a gm/100 sec.cm. The viscosity term is called dynamic viscosity

whereas kinematic viscosity is the dynamic viscosity divided by the density.

kinematic vis itydensity

cos dynamic viscosity

Kinematic viscosity has units of cm2/100 sec and the term is called centistoke.

Gas viscosity reduces as the pressure is decreased. At low pressures an increase in

temperature increases gas viscosity whereas at high pressures gas viscosity decreases

as the temperature increases. Figure 6 gives the values for pure component ethane.

1000900800700600

500

400

300

200

100908070

50 100 150 200 250 300 350 400

Temperature, deg F

Vis

co

so

ty,

mic

rop

ois

es

Viscosity of ethane

Pressure, psia5000

40003000

200015000

1000750

600

14.7

Figure 6 Viscosity of ethane

The viscosity of gases at low pressures can be obtained from correlations presented

by different workers.

Behaviour of Gases

20

50 100 150 200 250 300 350 400

Vis

cosity,

cp

Temperature, ?ºF

0.020

0.022

0.024

0.018

0.016

0.014

0.012

0.010

0.008

0.006

0.004

Helium

Nitrogen

Carbon D

ioxide

Methane

Ethylene

Ethane

Propane

i-Butane

n-Butane n-Pentane

n-Hexane

n-Heptane n-Octane

n-Nonane

n-Decane

Hydrogen Sulfide

Air

Figure 7 Viscosity of paraffin hydrocarbon gases at one atmosphere

Figure 7 and Figure 8 give the viscosities of individual components and paraffin

hydrocarbons at one atmosphere. For systems greater than 1 atmos the viscosities

can be obtained from the literature. Another way is by calculating the reduced

temperature and reduced pressure and use the chart developed by Carr6 which gives

a ratio of µ at reservoir conditions. This is given in Figure 9 in terms of pseudo

reduced conditions.


400 º F

300 º F

200 º F

100 º F

0.5 1.0 1.5 2.0 2.5 3.0 3.5

10 20 30 40 50 60 70 80 90 1000.004

0.005

0.006

0.007

0.008

0.009

0.010

0.011

0.012

0.013

0.014

0.015

0.016

Molecular Weight

Vis

cosity,

at 1 a

tm,

1, centipois

e

Gas Gravity (Air = 1)

N2

Mole per cent N2

G = 20

G = 06

G = 20

G = 06

1.5

1.0

1.5

1.0

G = 20

G = 06

1.5

1.0

0

0.0015

0.0010

0.0005

05 10 15

Co

rre

cti

on

ad

de

d t

o

Vis

co

sit

y,

c.p

.

0.0015

0.0010

0.0005

Co

rre

cti

on

ad

de

d t

o

Vis

co

sit

y,

c.p

.

0.0015

0.0010

0.0005

Co

rre

cti

on

ad

de

d t

o

Vis

co

sit

y,

c.p

.

CO2

Mole per cent CO2

00

5 10 15

H2S

Mole per cent H2S0

05 10 15

Figure 8 Viscosity of gases at atmospheric pressure6

Pseudoreduced Pressure, PR

0.8 1.0

1.0

1.5

2.0

2.5

3.0

3.5

4.0

5.0

6.0

1.2 1.4 1.6 1.8 2.0 2.2 2.4 2.6 2.8 3.0 3.2 3.4

= Viscosity at operating temperature

and pressure, centipoises

A = Viscosity at 14.7 psia (1atm) and

operating temperatures, centipoises

20

15

10

8

6

4

3

2

1

Pseudoreduced Temperature, TR

Vis

cosity,

/

A

Figure 9 Viscosity ratio vs pseudo reduced temperature and pseudo pressure.

Behaviour of Gases

22

5.2 Viscosity of MixturesAnother formula that is used for mixtures is:

mix

j j j

j j

y M

y M (30)

j = 1, n

where:

y = mole fraction of jth component

M = molecular weight of component

= the viscosity of jth component

n = number of components

j

j

j

The presence of other gases can also make a significant difference on the viscosity

(Figure 7).

EXERCISE 14.

Calculate the viscosity of the gas mixture in exercise 4 at 200°F and a pressure of

one atmosphere.

EXERCISE 15.

Use the gas gravity method to calculate the viscosity of the gas in exercise 4

EXERCISE 16.

Determine the viscosity of the gas in exercise 4 at 150°F and 3500 psia (ref ex 4, 7, &8)


6 EQUATIONS OF STATE

6.1 Other Equations-of-StateAs indicated at the start of section 2 the compressibility factor evolved out of the

need to use an equation derived out of ideal gas behaviour and incorporating it

into it a correction factor to suit real gas behaviour. One of the difficulties of the

compressibility equation:

PV = ZnRT

to describe the behaviour of gases is that the compressibility factor is not constant and

therefore mathematical manipulations cannot be made directly but must be carried out

through graphical or numerical techniques. Rather than use this modified equation

of state many have developed equations specifically to represent the behaviour of

real gases. It is an irony however that because of the long use of the equation above

incorporating z many of the real gas equation of states have been worked to calculate

z for use in the above equation.

6.2 Van de Waals Equation 1873The well known van der Waalʼs equation was one of the earliest equations to represent

the behaviour of real gases. This most basic EOS, which corrects for the volume of the

molecules and attractive and collision forces using empirical constraints a and b.

(P + a/V2) (V-b) = RT (31)

The two corrective terms to overcome the limiting assumptions of the ideal gas

equation are:

(i) The internal pressure or cohesion term , which accounts for the cohesion forces,

is a/V2.

(ii) The co-volume b, which represents the volume occupied by one mole at infinite

pressure and results from the repulsion forces which occur when the molecules

move close together.

The equation can also be written as:

V3 - (+ b) V2 + (a/P)V - ab/P = 0

Such equations are therefore called cubic equations of state.

The equation written to solve for z, the compressibility factor , becomes:

Z3 - Z2 (1 + B) + Z A - AB = 0 (32)

where

AaP

RTand B

bP

RT( )2

(33)

Behaviour of Gases

24

Values of a and b are positive constants for a particular fluid and when they are

zero the ideal gas equation is recovered. One can calculate P as a function of V for

various values of T. Figure 10 is a figure of 3 isotherms. Also drawn is the curve

for saturated liquid and saturated vapour.

Isotherm T1 is the single phase isotherm, T

c is the critical isotherm and T

2 gives the

isotherm below the critical temperature.

Vsat (liq) Vsat (vap)

V

P

Psat

c

T1>T

c

Tc

T2<T

c

Figure 10 PV behaviours of pure components predicted by EOS.

At the critical point , for a pure substance , the equation of state should be such

that:

P

V

P

VT T T Tc c

2

20

That is the critical isotherm exhibits a horizontal inflection point at the critical

point.


The application of these conditions to the van de Waals equation yields:

aR T

Pb

RT

P

c

c c

27

64 8

2 2

and

(34)

EXERCISE 17.

Calculate the critical constants for n- heptane.

For the curve, T2<Tc, the pressure decreases rapidly in the liquid region with increasing

V; after crossing the liquid saturated line a minimum occurs, rises to a maximum

and then decreases at the saturated vapour line. Real behaviour does not follow this

behaviour. They contain a horizontal segment where saturated liquid and saturated

vapour coexist in varying proportions.

This equation is not able to represent gas properties over a wide rage of temperatures

and pressures and over subsequent years many equations have been developed. A

number are given including those which are finding favour in their application in

this industry.

6.3 Benedict-Webb - Rubin Equation (BWR) 1940This equation developed for pure light hydrocarbons found considerable application

in predicting thermodynamic properties of natural gases, since natural gases are

essentially mixtures of light hydrocarbons and it can be written in a form similar to

Van der Waals equation.

PPT

V

B RT A C T

V

bRT a

V

a

V

C

V T V V

o o o

o

/

exp

2

2 3

6 3 2 2 21

(35)

where a, b, c, Ao, Bo and Co are constants for a given gas.

These equations are derived for pure components for which the empirical parameters

need to be obtained. For mixtures mixing rules are required to obtain these

constants.

6.4 Redlich-Kwong Equation 1949Numerous equations were developed with increasing numbers of constants specific to

pure components. More recently there has been a move back to the cubic equations

like van der Waals. We will describe briefly those which have found favour in the

oil and gas sector.

This modern development of cubic equations of state started in 1949 with the Redlich

and Kwong equation which involves only two empirical constants.

Behaviour of Gases

26

P = RT

V - b

a T

V V b (36)

where a and b are functions of temperature.

The term a(T) depends on the temperature and Redlich Kwong expressed this as a

function of the reduced temperature Tr using

a Ta

T

c

R

By applying the limiting condition at the critical points yields values of ac and b

related to critical constants. Such that ;

aR T

Pb

RT

Pc

c

c

c

c

0 42748 0 086642 2

. . and

(37)

6.5 Soave, Redlich Kwong equationSoave, in 1972, modified the Redlick-Kwong (RK) equation and replaced the a/T0.5

term with a temperature dependent term aT where a

T = a

c. .

The Soave, Redlich-Kwong (SRK) equation is therefore:

PRT

V b

a

V V b

c

(38)

where

is a non dimensionless temperature dependent term which has a value of 1.0 at the

critical temperature.

is obtained from

1 12

2

m Tr

where m = 0.480 + 1.574 - 0.176

where is the Pitzer accentric factor .8

6.6 Peng Robinson Equation of State 1975Peng and Robinson modified previous equations in relation to the attractive term.

They introduced it to improve the predictions of the Soave modification in particular

for the calculation of liquid densities.


PRT

V b

a

V V b b V b

c

(39)

aR T

Pb

RT

Pc

c

c

c

c

0 457235 0 07782 2

. . and

(40)

and

is the same function as for the Soave equation except the function is

different;

where m = 0.37464 + 1.54226w - 0.26992w2

These equations, in particular the SRK and PR equation are widely used in simulation

software used to predict behaviour in reservoirs, wells and processing. There are

other equations of state which are as competent at predicting physical properties

which have been developed mainly focusing on the need to improve the accuracy of

liquid volumes predictions. There is, however, great reluctance to change from those

presently used because of the investment in their associated parameters. An excellent

review of these equations and application is given by Danesh 9.

6.7 Application to MixturesWhen properties of mixtures are required mixing rules are required to combine the

data from pure components.

For both the SRK and PR equation

b y b y y a aj kj j

j

i j i

ji

ij and a 1

(41)

where the term kij is termed the binary interaction coefficients which are independent

of pressure and temperature. Values of binary interaction coefficients are obtained

by fitting equation of state (EOS) predictions to gas-liquid data for binary mixtures.

They have NO physical property significance. Each equation has its own binary

interaction coefficient.

Effort is underway and methods exist to not use binary interaction parameters but to

use physical property related parameters to enable good quality predictions.

Behaviour of Gases

28

EXERCISE 18.

A PVT cell contains 0.01 cu ft ( 300cc) of gas with at composition of ; methane 0.67 mol.frac, ethane 0.235 and n-butane 0.05. The temperature is increased to 300°C. Use the SRK equation to calculate the pressure at this increased temperature. Use

binary interaction coefficients of C1-nC4 0.02, C2-nC4 0.01 and C1-C2 0.0

Solutions to Exercises

EXERCISE 1.

A gas cylinder contains methane at 1000 psia and 70 oF. If the cylinder has a volume

of 3 cu.ft assuming methane is an ideal gas calculate the mass of methane in the

cylinder.

SOLUTION

PV = nRT

n = m/M

where n = number of moles

m = mass

M = molecular weight

m = PMV/RT

m

psialb

lbmolecuft

psia cuft

lbmole RR

o

o

1000 16 04 3

10 73 530

.

..

.

Mass of methane, m = 8.46 lb


EXERCISE 2.

Calculate the density of the gas in the cylinder in exercise 1.

SOLUTION

g = MW.P

RT

g

g

psialb

lbmolepsia cuft

lbmole oRR

Density of gaslb

cu ft

1000 16 04

10 73 530

2 82

0

.

..

.

, .. .

EXERCISE 3.

Assuming methane is at the conditions of exercise 1, calculate the volume the gas

would occupy at standard conditions.

SOLUTION

P V

T =

P V

T

P V

T

= P

P T

T V

= 1000

1 520 R

530 R x3ft

= 200.23 scf

1 1

1

2 2

2

sc sc

sc

1

sc

sc

1

psia

psia

o

o

3

V

V

V

sc

sc

sc

4 7.

EXERCISE 4.

A gas is made up of the following components; 25lb of methane, 3 lb of ethane and 1.5

lb of propane. Express the composition of the gas in weight and mole fractions.

Behaviour of Gases

30

SOLUTION

Gas Components

AWeight

B Mol weight

Clb moles

DMole fraction

Methane 25 16.04 1.559 0.921

Ethane 3 30.07 0.100 0.059

Propane 1.5 44.09 0.034 0.020

Totals 29.05 1

1

2

3

EXERCISE 5.

What is the apparent molecular weight of the gas in exercise 4

SOLUTION

Apparent Molecular weight= 17.43

Gas Components

AMol weight

BMol fraction

C

Methane 16.04 0.921 14.77

Ethane 30.07 0.059 1.77

Propane 44.09 0.020 0.89

1.000 17.43

1

2

3

mw yi A*B

EXERCISE 6.

What is the gas gravity of the gas in exercise 4 ?

SOLUTION

g

g

air

gM

M

M

29

g = AMW = 17.43

Gas gravity = 0.6

EXERCISE 7.

Calculate the pseudo critical temperature and pseudocritical pressure of the mixture

in exercise 4 .


SOLUTION

1

2

3

Gas Components

AMol weight mw

BMole fraction yi

C Pc-psi

D.

Tc °R Ppc

Methane 16.04 0.921 667.00 344 614.3 316.81

Ethane 30.07 0.059 708.00 550 41.7 32.42

Propane 44.09 0.020 616.00 666 12.4 13.39

Total 1.0 668.4 362.6

Tpc

Pseudocritical pressure = 668.4 psia

Pseudocritical temperature = 362 oR

EXERCISE 8.

For the gas of exercise 4 determine the compressibility factor at a temperature of 150 oF and a pressure of 3500psia.

SOLUTION

Ppr = P/P

pc, T

pr = T/T

pc

From exercise 6 Ppc

= 668 psia, Tpc

= 362.6°R

P = 3500 psia, and T = 150°C ie. 610°R

Ppr = 5.24, and Tpr = 1.68

From standing Katz chart, figure 2

Compressibility factor, z = 0.88

EXERCISE 9.

Calculate the pseudo critical properties of the gas in exercise 4 if it also contained 3

lb of hydrogen sulphide, 10lb of carbon dioxide and 2.5lb of nitrogen

1

2

3

Gas

Components

Mol

weight

Mole

fraction

Pc-psi Tc °R Ppcpsia

Methane 25 0.56 16.04 0.035 0.743 667.00 344 495.8 255.70

Ethane 3 0.07 30.07 0.002 0.048 708.00 550 33.7 26.17

Propane 1.5 0.03 44.09 0.001 0.016 616.00 666 10.0 10.81

Hydrogen 3 0.07 34.08 0.002 0.042 1306 673 54.8 28.25

sulphide

Carbon 10 0.22 44.01 0.005 0.108 1071 548 116.1 59.38

Dioxide

Nitrigen 2.5 0.06 28.02 0.002 0.043 493 227 21.0 9.66

Total 45 1.00 0.0466 1.000 731 390

TpcWeight Wgt

fraction

lb moles

4

5

6

From Wichert & Azis chart for compositions of H2S and CO

2 = 19

Behaviour of Gases

32

T = T - = 371 R

P = 694.3

pc pc

o

pc

pp T

T y ypc

pc pc

pc H S H S2 21

EXERCISE 10.

Express the quantity of 1 lb mole of a gas as standard cubic feet.

SOLUTION

Equation of state PV = RT for 1 mole

R = 10.732 psia. cu.ft/lb.mole °R T = 60+460 = 520 °R, P = 14.65 psia

or V for 1 lb.mole = RT/P = 380.9 scf/lb.mole.

EXERCISE 11.

Express the mass of gas in exercise 4 as standard cubic feet.

SOLUTION

Total mass of gas = 29.5 lb.

Apparent mol.wgt of gas exercise 5 = 17.43 lb./lb.mole

lb.moles of gas = 1.6924

Standard cubic feet of gas = 380.9 x 1.6924

= 644.68 scf

EXERCISE 12.

Calculate the gas formation factor for a gas with the composition of exercise 4 existing

at the reservoir conditions given in exercise 8.

SOLUTION

T = 150 oF ie 610 oR and P = 3500 psia

Compressibility factor at these conditions from exercise 8 = 0.88

Bg using equation above = 0.0008 res bbl/scf


EXERCISE 13.

A reservoir exists at a temperature of 150oF (as for exercise 8) suitable for storing

gas. It has an areal size of 5 miles by 2 miles and is 200ft thick. The average porosity

is 20% and there is no water present. How much gas of the composition of exercise

4 can be stored at a pressure the same as in exercise 8 i.e. 3500 psia. ? (1 mile=

5280 ft.)

SOLUTION

Volume of reservoir pore space = 5x2 x (5280)2 x 200 x 0.2

= 11,151,360,000 cu. ft.

=1,985,994,657 bbls

Bg , exercise 11 =0.00077299 res. bbls/SCF

Volume of gas =2.56923E+12 scf

EXERCISE 14.

Calculate the viscosity of the gas mixture in exercise 4 at 200°F and a pressure of

one atmosphere.

SOLUTION

Gas

Components

Mol Weight Mole fraction

yj

Viscosity

from fig 7

j

Mj yj Mj

Methane

Ethane

Propane

jyj Mj

16.04

30.07

44.09

0.921

0.059

0.020

1.000

0.013

0.0112

0.0098

4.0050

5.4836

6.6400

SUM

3.6884

0.3233

0.1335

4.1451

0.0470

0.0036

0.0013

0.529

mix

j j j

j j

y M

y M

mix = 0.0529/4.1451

mix =0.01275 cp

EXERCISE 15.

Use the gas gravity method to calculate the viscosity of the gas in exercise 4

SOLUTION

Gas Components

Mol Weightmw

Mole fractionyj

Methane

Ethane

Propane

16.040

30.070

44.090

0.000

0.921

0.059

0.020

1.000

14.7720

1.773

0.886

17.431

Behaviour of Gases

34

g=AMW/M

air

g=AMW/29 Temperature = 150°F

Mol weight air = 29.000

AMW of gas = 17.431

Gas Gravity = 0.601

g = 0.01265 from fig 8

EXERCISE 16.

Determine the viscosity of the gas in exercise 4 at 150oF and 3500 psia (ref ex 4, 7,

&8)

SOLUTION

From exercise 7

P

T

P

TT

pc

pc

r

rc

668 4

362 6

.

.

PP =

3500

668.4 = 5.24

T = 610

362.6 = 1.68

c

From Lee correlation

/ atmos = 1.75

Viscosity at atmospheric pressure

From exercise 13 and 14 = 0.01275 cp

Viscosity at conditions = 0.0223 cp

EXERCISE 17.

Calculate the critical constants for n- heptane.

SOLUTION

R = 10.732. Tc for heptane = 973 oR and P

c = 397 psia

Using equations above a = 115,872 cu ft 2/lb mole

and b = 3.2878 cu ft./lb mole


EXERCISE 18.

A PVT cell of volume 0.01 cu ft ( 300cc) contains 0.008 lb mole. of gas with

a composition of; methane 0.67 mol.frac, ethane 0.235 and n-butane 0.05. The

temperature is increased to 300°C. Use the SRK equation to calculate the pressure

at this increased temperature. Use binary interaction coefficients of C1-nC4 0.02,

C2-nC4 0.01 and C1-C2 0.0

SOLUTION

Calculate the constants a and b for each component

aR T

Pb

RT

P

m T

cc

c

c

c

r

0 42748 0 08664

1 1

2 2

2

. . and

where m = 0.480 + 1.574 - 0.176 2

Components Pc ac m a a=a*Y Tc°R) bj

Methane

ethane

n-butane

0.67 344 667 0.4759 8735 0.0104 0.49635 0.57546 5027

0.235

0.05

550

766

708

551

0.7223

1.2926

21036

52429

0.0979

0.1995

0.63241

0.78701

0.79033

1.00619

16625

52753

Now calculate the mixture values.

b y b y y a aj kj j

j

i j i

ji

ij and a

where a = (1- k )(a a )ij ij i j

0.5

1

Components kijMethane

kijn-butane

aijMethane

aijethane

aijn-butane

sumYi b kijethane

Methane

ethane

n-butane

0.67 0.312 0.00 0.00 0.02 2123.7 1485.5 1037.29 4646.52

0.235

0.05

0.181

0.129

0.00 0.01

0.00

1485.5

1037.3

1039.1

732.97

732.969

527.535

3257.56

2297.8

1 0.622 sum 10201.9

Now use SRK to calculate pressure.

Behaviour of Gases

36

PRT

V b

a

V V b

c

V = 1.25 cu ft / lb mole

b = 0.622 a = 10201.9

P = 8617.6 psia

m

c

REFERENCES

1. Standing MB and Katz DL Density of Natural Gases. Trans AIME, 146(1942).

p140

2. Poettmann FH and Carpenter PG The Multiphase Flow of Gas and Water through

Vertical Flow Strings with Application to the Design of Gas Lift Installations.

API Drilling and Production Practise. 1952, pp 279-91

3. Brown GG et al. Natural Gasoline and Volatile Hydrocarbons” National Gasoline

Assoc. of America, Tulsa, Okl. 1948

4. Wichert, E and Aziz,K “ Calculate Zʼs for sour gases” Hyd Proc.(May 1972)

51, 119-122

5. Katz, D.L., Handbook of Natural Gas Engineering, McGraw Hill, NY, 1959

6. Carr N et al. Viscosity of natural gases under pressure. Trans AIME 201, 264,

(1954)

7. Lee et al “The viscosity of natural gases.” Trans AIME 1966 237, 997-1000

8. Pitzer K S et al The Volumetric and Thermodynamic Properties of Fluids II.

Compressibility Factor, Vapour Pressure and Entropy of Vaporisation. J .Am.

Chem. Soc. (1955) 77, No 13,3433-3440

9. Danesh, A PVT and Phase Behaviour of Petroleum Reservoir Fluids. 1998

Elsevier ISBN:0 444 82196 1 p129-162

CONTENTS

1 COMPOSITION BLACK OIL MODELS

2 GAS SOLUBILITY, Rs

3 OIL FORMATION VOLUME FACTOR, Bo

4 TOTAL FORMATION VOLUME FACTOR, BT

5 BELOW THE BUBBLE POINT

6 OIL COMPRESSIBILITY

7 BLACK OIL CORRELATIONS

8 FLUID DENSITY

8.1 Specific Gravity of a Liquid

8.2 Density Based on Ideal Solution Principles

9 FORMATION VOLUME FACTOR OF GAS

CONDENSATE, Bgc

10 VISCOSITY OF OIL

11 INTERFACIAL TENSION

12 COMPARISON OF RESERVOIR FLUID

MODELS

Properties of Reservoir Liquids

2

LEARNING OBJECTIVES

Having worked through this chapter the Student will be able to:

• Define gas solubility, Rs and plot vs. P for a reservoir fluid.

• Define undersaturated and saturated oil.

• Explain briefly flash and differential liberation

• Define the oil formation volume factor Bo, and plot B

o vs. P for a reservoir

fluid.

• Define the Total Formation Volume factor Bt, and plot B

t vs. P alongside a Bo

vs. P plot.

• Present an equation to express Bt in terms of B

o, R

s and B

g.

• Express oil compressibility in terms of oil formation volume factor.

• Use black oil correlations and their graphical form to calculate fluid

properties.

• Calculate the density of a reservoir fluid mixture, using ideal solution principles,

at reservoir pressure and temperature, using density correction chart for C1 &

C2 and other prerequisite data.

• Define the formation volume factor of a gas condensate

• Calculate the reserves and production of gas and condensate operating above

the dewpoint, given prerequisite data.

• Use viscosity equations and correlations to calculate viscosity of fluid at reservoir

conditions.

• Calculate the interfacial tension of equilibrium gas-oil systems given prerequisite

equations and data.

• List the comparisons of the black oil and compositional model in predicting

liquid properties


1 COMPOSITION - BLACK OIL MODEL

As introduced in the chapter on Composition, petroleum engineers are requiring

a compositional description tool to use as a basis for predicting reservoir and well

fluid behaviour. The two approaches that are commonly used are the multicomponent

compositional model described in the earlier chapter and the two component black oil

model. The latter simplistic approach has been used for many years to describe the

composition and behaviour of reservoir fluids. It is called the “Black Oil Model”.

The black oil model considers the fluid being made up of two components - gas dissolved

in oil and stock tank oil. The compositional changes in the gas when changing pressure

and temperature are ignored. To those appreciating thermodynamics this simplistic

two component model is difficult to cope with. The Black Oil Model, illustrated in

Figure 1, is at the core of many petroleum engineering calculations, and associated

procedures and reports.

Associated with the black oil model are Black Oil model definitions in relation to

Gas Solubility and Formation Volume Factors.

Reservoir Fluid

Solution Gas

Stock Tank Oil

/ = Rs

/ = Bo

Bo = Oil Formation Volume Factor

Rs = Solution Gas to Oil Ratio

Figure 1 "Black Oil Model"


4

2 GAS SOLUBILITY

Although the gas associated with oil and the oil itself are multicomponent mixtures

it is convenient to refer to the solubility of gas in crude oil as if we were dealing with

a two-component system.

The amount of gas forming molecules in the liquid phase is limited only by the

reservoir conditions of temperature and pressure and the quantity of light components

present.

The solubility is referred to some basis and it is customary to use the stock tank

barrel.

Solubility = f (pressure, temperature, composition of gas

composition of crude oil)

For a fixed gas and crude, at constant T, the quantity of solution gas increases with

p, and at constant p, the quantity of solution gas decreases with T

Rather than determine the amount of gas which will dissolve in a certain amount of

oil it is customary to measure the amount of gas which will come out of solution as

the pressure decreases. Figure 2 illustrates the behaviour of an oil operating outside

the PT phase diagram in its single phase state when the reservoir pressure is above

its reservoir bubble point at 1. Fluid behaviour in the reservoir is single phase and

the oil is said to be undersaturated . In this case a slight reduction of pressure causes

the fluid to remain single phase. If the oil was on the boundary bubble point pressure

line at 2 then a further reduction in pressure would cause two phases to be produced,

gas and liquid. This saturated fluid is one that upon a slight reduction of pressure

some gas is released. The concept of gas being produced or coming out of solution

gives rise to this gas solubility perspective. Clearly when the fluids are produced to

the surface as shown by the undersaturated oil in figure 2 the surface conditions lie

within the two phase area and gas and oil are produced. The gas produced is termed

solution gas and the oil at surface conditions stock tank oil. These are the two com-

ponents making up the reservoir fluid, clearly a very simplistic concept.

The gas solubility Rs is defined as the number of cubic feet (cubic metre) of gas

measured at standard conditions, which will dissolve in one barrel (cubic metre)

of stock tank oil when subjected to reservoir pressure and temperature.

In metric units the volumes are expressed as cubic metre of gas at standard conditions

which will dissolve in one cubic metre of stock tank oil.


Solution Gas

Stock Oil Tank

Oil Reservoir

Oil and Dissolved Gas

Rsi scf/stb

1 st b. oil

Bo rb.oil

Pre

ssure

Temperature

Pi 1

2

P

+

Surface

Phase Diagram

Figure 2 Production of reservoir hydrocarbons above bubble point

Figure 3 gives a typical shape of gas solubility as a function of pressure for a reser-

voir fluid at reservoir temperature. When the reservoir pressure is above the bubble

point pressure then the oil is undersaturated, i.e. capable of containing more gas. As

the reservoir pressure drops gas does not come out of solution until the bubble point

is reached, over this pressure range therefore the gas in solution is constant. At the

bubble point pressure, corresponding to the reservoir temperature, two phases are

produced, gas and oil. The gas remaining in solution therefore decreases.

The nature of the liberation of the gas is not straight forward. Within the reservoir

when gas is released then its transport and that of the liquid is influenced by the relative

permeability of the rock ( discussed in Chapter 10). The gas does not remain with its

associated oil i.e. the system changes. In the production tubing and in the separator

it is considered that the gas and associated liquid remain together i.e. the system is

constant. The amount of gas liberated from a sample of reservoir oil depends on the

conditions of the liberation. There are two basic liberation mechanisms:


6

1000 2000 3000

200

600

400

Pressure (psig)

Pb

Rsi

Rs scf/stb

Figure 3 Solution Gas - Oil Ratio as a Function of Pressure.

Flash liberation - the gas is evolved during a definite reduction in

pressure and the gas is kept in contact with the liquid

until equilibrium has been established.

Differential liberation - the gas being evolved is being continuously

removed from contact with the liquid and the liquid is in

equilibrium with the gas being evolved over a finite

pressure range.

The two methods of liberation give different results for Rs. This topic is covered in

more detail in the PVT analysis chapter.

Production of a crude oil at reservoir pressures below the bubble point pressure occurs

by a process which is neither flash or differential vaporisation. Once enough gas is

present for the gas to move toward the wellbore the gas tends to move faster than the

oil. The gas formed in a particular pore tends to leave the liquid from which it was

formed thus approximating differential vaporisation, however, the gas is in contact with

liquid throughout the path through the reservoir. The gas will also migrate vertically

as a result of its lower density than the oil and could form a secondary gas cap.

Fluid produced from reservoir to the surface is considered to undergo a flash process

where the system remains constant.

3 OIL FORMATION VOLUME FACTOR, B o

The volume occupied by the oil between surface conditions and reservoir or other

operating changes is that of the total system; the ʻstock tank oil ̓ plus its associated

or dissolved ʻsolution gasʼ. The effect of pressure on the complex stock tank liquid

and the solution gas is to induce solution of the gas in the liquid until equilibrium is

reached. A unit volume of stock tank oil brought to equilibrium with its associated


gas at reservoir pressure and temperature will occupy a volume greater than unity

(unless the oil has very little dissolved gas at very high pressure).

The relationship between the volume of the oil and its dissolved gas at reservoir

condition to the volume at stock tank conditions is called the Oil Formation Volume

Factor Bo. The shape of the B

o vs. pressure curve is shown in Figure 4. It shows

that above the bubble point pressure the reduction in pressure from the initial pres-

sure causes the fluid to expand as a result of its compressibility. This relates to the

chapter on Phase Behaviour where for an oil the PV diagram shows a large decline

in pressure for a small increase in volume, being again an indication of the com-

pressibility of the liquid. Below the bubble point pressure this expansion due to

compressibility of the liquid is small compared to the ʻshrinkage ̓of the oil as gas is

released from solution.

The oil formation volume factor, is the volume in barrels (cubic metres) occupied in

the reservoir, at the prevailing pressure and temperature, by one stock tank barrel

(one stock tank cubic metre) of oil plus its dissolved gas.

1000 2000 30001.0

1.2

1.1

Pressure (psig)

Pb

Bo

rb

./stb

Units - rb (oil and dissolved gas)

Figure 4 Oil formation volume factor

These black oil parameters, Bo and Rs are illustrated in Figure 5 a,b,&c from Craft

and Hawkins 1 reservoir engineering text., where they present the Rs and B

o curve for

the Big Sandy field in the USA. The visual concept of the changes during pressure

and temperature decrease is also presented.


8

P01

P01

= 3500 PSIA

T01

= 160º F

A

PB

= 2500 PSIA

T01

= 160º F

B

P = 1200 PSIA

T01

= 160º F

C

PA

= 14.7 PSIA

T01

= 160º F

D

PA

= 14.7 PSIA

T01

= 60º F

E

PB

P

PA PA

Free Gas

676 Cu. Ft.Free Gas

2.990 Cu. Ft.

Free Gas

567 Cu. Ft.

1,000 BBL1,040 BBL1,210 BBL1,333 BBL1,310 BBL

567SCF/STB

AT 1200 PSIA

RS = 337

BU

BB

LE

PO

INT

PR

ES

SU

RE

INIT

IAL P

RE

SS

UR

E

Solu

tion G

as, S

CF

/ST

B

600

500

400

300

200

100

00 500 1000 1500 2000

Pressure, PSIA

2500 3000 3500

(a)

(b)

Figure 5 Gas to oil ratio and oil formation volume factor for Big Sandy Field reservoir

oil 1.

Form

ation V

olu

me F

acto

r, B

BL/S

TB

0 500

1.40

1.30

1.20

1.10

1.001000 1500 2000

Pressure, PSIA

2500 3000 3500

BU

BB

LE

PO

INT

PR

ES

SU

RE

INIT

IAL P

RE

SS

UR

E1200 PSIA

BO

= 1.210

14.7 PSIA & 160º F

BO

= 1.040

2500 PSIA

BOB

= 1.333

3500 PSIA

BOI

= 1.310

14.7 PSIA & 60º F

BO

= 1.000

(b)

Figure 5b


The reciprocal of the oil formation volume factor is called the ʻshrinkage factor bo

bB

o

o

1

The formation factor Bo may be multiplied by the volume of stock tank oil to find

the volume of reservoir required to produce that volume of stock tank oil. The

shrinkage factor can be multiplied by the volume of reservoir oil to find the stock

tank volume.

It is important to note that the method of processing the fluids will have an effect

on the amount of gas released and therefore both the values of the solution gas-oil

ratio and the formation volume factor. A reservoir fluid does not have single Bo or R

s

values. Bo & R

s are dependant on the surface processing conditions. This simplistic

reservoir model (Figure 6) demonstrates that the black oil model description of the

reservoir fluids is an after the event, processing, description in terms of the produced

fluids. This simplistic approach to modelling reservoir fluids becomes more difficult

to consider when one is involved in reservoirs which become part of a total reservoir

system (Figure 7).

Rs

BO

Figure 6 Black oil description of reservoir fluid


10

Rs 3

Bo 3

Rs 2

Bo 2

Rs 4

Bo 4

Rs 1

Rs

Bo

Bo 1

?

Multi Reservoir System

Figure 7 Integrated system of reservoir common pipeline and final collection system.

4 TOTAL FORMATION VOLUME FACTOR, Bt

In reservoir engineering it is sometimes convenient to know the volume occupied

in the reservoir by one stock tank barrel of oil plus the free gas that was originally

dissolved in it. A factor is used called the total formation-volume factor Bt, or

the two-phase volume-factor and is defined as the volume in barrels that 1.0 STB

and its initial complement of dissolved gas occupies at reservoir temperature and

pressure, i.e. it includes the volume of the gas which has evolved from the liquid

and is represented by:

Bg (R

sb - R

s)

i.e. Bt = B

o + B

g (R

sb - R

s) (1)

Rsb

= the solution gas to oil ratio at the bubble point


Oil

Oil

Gas

Hg

B0

Bt

B0b

Bg(Rsb-Rs)

Figure 8a Total formation volume factor or two phase volume factor

Its application comes from the Material Balance equation (Chapter 15) where it is

sometimes used to express the volume of oil and associated gas as a function of pres-

sure. It is important to note that Bt does not have volume significance in reservoir

terms since the assumption in Bt is that the system remains constant. As mentioned

earlier if the pressure drops below the bubble point in the reservoir then the gas

coming out of solution moves away from its associated oil because of its favourable

relative permeability characteristics.

Figure 8b gives a comparison of the total formation-volume factor with the oil for-

mation-volume factor. Clearly above Pb the two values are identical since no free

gas is released. Below Pb the difference between the values represents the volume

occupied by free gas.

BoBt

Pressure Pb

Figure 8b Total and oil formation volume factor

The value of BT can be estimated by combining estimates of B

O and calculation of

Bg and known solubility values for the pressures concerned.


12

5 BELOW THE BUBBLE POINT

Figure 9 depicts the behaviour below the bubble point when produced gas at the

surface comes from two sources, the solution gas associated with the oil entering the

wellbore plus free gas which has come out of solution in the reservoir and migrated to

the wellbore. The total producing gas to oil ratio is made up of the two components

solution gas Rs and the free gas which is the difference. The diagram illustrates the

volumes occupied by these two in the reservoir, the solution gas being part of Bo and

the free gas volume through Bg.

Free Gas

& Solution Gas

Stock Oil Tank

Oil Reservoir

rb (oil and dissolved gas) /stb

1 st b. oil

Bo

Pre

ssure

Temperature

R= Rs + (R - Rs)

+

(R - Rs) Bg

Gas Oil

Reservoir

rb (free gas) /stb

SurfacePi

P

Figure 9 Production of reservoir hydrocarbons below bubble point

6 OIL COMPRESSIBILITY

The volume changes of oil above the bubble point are very significant in the context

of recovery of undersaturated oil. The oil formation volume factor variations above

the bubble point reflect these changes but they are more fundamentally embodied

in the coefficient of compressibility of the oil, or oil compressibility.

The equation for oil compressibility is

cV

V

Po

T

1

in terms of formation volume factors this equation yields


cB

B

Po

o

o

T

1

Assuming that the compressibility does not change with pressure the above equation

can be integrated to yield ;

c P PV

Vo 2 1

2

1

ln

where P1 & P

2, and V

1 & V

2 represent the pressure and volume at conditions 1 & 2.

7 BLACK OIL CORRELATIONS

Over the years there have been many correlations generated based on the two com-

ponent based black oil model characterisation of oil. The correlations are based

on data measured on the oils of interest. These empirical correlations relate black

oil parameters, the variables of Bo and R

s to; reservoir temperature, and oil and gas

surface density. It is important to appreciate that these correlations are empirical and

are obtained by taking a group of data for a particular set of oils and finding a best fit

correlation. Using the correlation for fluids whose properties do not fall within those

for the correlation can result in significant errors. Danesh 2 has given an excellent

review of many of these correlations

A number of empirical correlations, based on largely US crude oils, and other loca-

tions across the world have been presented to estimate black oil parameters of gas

solubility and oil formation volume factor. The most commonly used is Standingʼs 3 correlation. Other correlations include, Lasater 4, and recently Glaso 6

Pb = f (R

s,

g, p

o, T)

where Pb = bubble point pressure at ToF

Rs = solution gas-oil ratio (cu ft/ bbl)

g= gravity of dissolved gas

o = density of stock-tank oil .(specific gravity)

Standingʼs correlation for the calculation of Pb, bubble point pressure is:

PR

T APIbs

g

. ( . . ( )) .

.

18 2 0 00091 0 0125 1 4

0 83

10

(2)

His correlation for the oil formation volume factor is;

B R To s

g

o

. . .

. .

0 9759 0 000120 1 25

0 5 1 2

(3)


14

Standing's correlations have been presented as nomographs enabling quick look

predictions to be made. Figures 10 & 11 give the nomogram forms of these correlations

for gas solubility and oil formation volume factor. Standingʼs correlation is based on

a set of 22 California crudes.

Other correlations have been presented by Lasater 4 based on 137 Canadian ,USA

and South American crudes, Vasquez and Beggs5 using 6000 data points, Glaso 6 us-

ing 45 North Sea crude samples, and Mahoun 7 who used 69 Middle Eastern crudes.

Danesh2 gives a very useful table showing the ranges covered by the respective black

oil correlations


Figure 10 Oil-formation volume factor as a function of gas solubility, temperature, gas

gravity and oil gravity (Standing)

20

30

40

50

60

70

8090100

150

200

300

400

500

600

700800

9001000

1500

2000

1.021.03

1.041.05

1.061.07

1.081.09

1.10

Formation volume of bubble-point liquid

Gas

-oil

ratio

, cu ft

per

bbl

bbl p

er b

bl o

f ta

nk

oil

1.20

1.30

1.40

1.50

1.60

1.70

1.80

1.90

1.10

1.20

1.30

1.40

1.50

0.5

0 0.6

0 0.7

0 0.8

0 0.9

0 1.0

0

Ga

s g

rav

ity

Air

=1

Tank oil gravity, ºAPI50 30 10

Temperature, ºF

100

140160

180200

220240

260

120


16

Fig

ure 11

Gas so

lub

ility as a fu

nctio

n o

f pressu

re. Tem

peratu

re, gas g

ravity

and

oil

grav

ity

600

500

400

300

200

20

30

40

50

60

7080

90100

150

200

300

400

500

600

700

700

800

900

1000

1500

2000

3000

4000

5000

6000

800900

1000

1500

2000

Tank oil gravity, ºAPI

Temperature, ºF

Gas gravity Air = 1

60

1.501.40

1.301.20

80

100

120

140

160

180

200

220240

260

1.101.000.900.80

1014

1618

2022

2426

2830

3234

3638

40

12

4244

4648

5052

5456

58

Bu

bb

le-p

oin

tPressure, psia

Gas-oil ratio, cu ft per bbl

60

(STA

ND

ING

)

0.700.600.50


Correlation Standing Lasater Vasquez-Beggs Glaso Marhoun

Ref 3 4 5 6 7

Bubble - point pressure (psia) 130-7000 45-5780 15-6055 165-7142 130-3573

Temperature, F 100-258 82-272 162-180 80-280 74-240

Bo 1.024-2.15 1.028-2.226 1.025-2.588 1.032-1.997

Gas - oil ratio (scf/stb) 20-1425 3-2905 0-2199 90-2637 26-1602

Oil Gravity, oAPI 16.5-63.8 17.9-51.1 15.3-59.5 22.3-48.1 19.4-44.6

Gas Gravity 0.59-0.95 0.574-1.22 0.511-1.651 0.65-1.276 0.752-1.367

Separator Pressure 265-465 15-605 60-565 415

Searator Temperature F 100 36-106 76-150 125

Table 1 Black oil correlation and their ranges at application 2

8 FLUID DENSITY

Liquids have a much greater density and viscosity than gases, and the density is affected

much less by changes in temperature and pressure. For petroleum engineers it is

important that they are able to estimate the density of a reservoir liquid at reservoir

conditions.

8.1 Specific Gravity of a Liquid

o

o

w (4)

The specific gravity of a liquid is the ratio of its density to that of water both at the

same T & P. It is sometimes given as 60˚/60˚, i.e. both liquid and water are measured

at 60˚ and 1 atmos.

The petroleum industry uses another term called ˚API gravity where

APIo

141 5131 5

..

(5)

where o is specific gravity at 60˚/60˚.

There are several methods of estimating the density of a petroleum liquid at reservoir

conditions. The methods used depend on the availability and nature of the data of

data. When there is compositional information on the reservoir fluid then the density

can be determined using the ideal solution principle. When the information we have

is that of the produced oil and gas then empirical methods can be used to calculate

the density of the reservoir fluid.

8.2 Density based on Ideal Solution PrinciplesMixtures of liquid hydrocarbons at atmospheric conditions behave as ideal solutions.

An ideal solution is a hypothetical liquid where no change in the character of the

liquids is caused by mixing and the properties of the mixture are strictly additive.


18

Petroleum liquid mixtures are such that ideal-solution principles can be applied for the

calculation of densities and this enables the volume of a mixture from the composi-

tion and the density of the individual components. The principle is illustrated using

the following exercise. Data for the specific components are given in the tables at

the end of the chapter

EXERCISE 1.

Calculate the density at 14.7psia and 60 ºF of the hydrocarbon liquid mixture with the composition given below:

Component Mol.

fract.

1b mol.

nC4 0.25

nC5 0.32

nC6 0.43

1.00

SOLUTION EXERCISE 1

Solution Component Mol. Mol. Weight Liquid Liquid

density

fract. weight 1b Density at volume

1b mol. 1b/1b at 60˚F and 14.7 cu ft

mol. psia

1b/cu ft

nC4 0.25 58.1 14.525 36.45 0.3985

nC5 0.32 72.2 23.104 39.36 0.5870

nC6 0.43 86.2 37.066 41.43 0.8947

____ _____ _____

1 74.695 1.8801

Liquids at their bubble point or saturation pressure contain large quantities of dis-

solved gas which at surface conditions are gases and therefore some consideration

for these must be given in the additive volume technique. This physical limitation

does not impair the mathematical use of a “pseudo liquid density “ for methane and

ethane since it is only a step in its application to determine a reservoir condition

density. This is achieved by obtaining apparent liquid densities for these gases and

determining a pseudoliquid density for the mixture at standard conditions which can

then be adjusted to reservoir conditions.

Standing & Katz 8 carried out experiments on mixtures containing methane plus other

compounds and ethane plus other compounds and from this were able to determine

a pseudo-liquid (fictitious) density for methane and ethane


Correlations have been obtained by experiment giving apparent liquid densities of

methane and ethane versus the pseudoliquid density (Figure 12).

0.1

0.2

0.3

0.4

0.5 0.6 0.7 0.8 0.9

0.3

0.4

0.5

0.6

0.40.3

Density of system, 60ºF B atm. pressure

Ap

pa

rre

nt

de

ns

ity

of

Me

tha

ne

, g

/cc

Ap

parr

en

t d

en

sit

y o

f o

f E

than

e, g

/cc

Ethane - N - Butane

Ethane - Heptane

Ethane - Crystal oil

Methane - Cyclo Hexane

Methane - Crude oil

Methane - Crystal oil

Methane - Propane

Methane - Hexane

Methane - Pentane

Methane - Heptane

Methane - Benzene

Figure 12 Variation of apparent density of methane and ethane with density of the system 8.

To use the correlations a trial and error technique is required whereby the density of

the system is assumed and the apparent liquid densities can be determined. These

liquid densities are then used to compute the density of the mixture by additive vol-

umes and the value checked against the initial assumption. The procedure continues

until the two values are the same.

When non hydrocarbons are present, the procedure is to add the mole fractions of

the nitrogen to methane, the mole fraction of carbon dioxide to ethane and the mole

fraction of hydrogen sulphide to propane.


20

EXERCISE 2:

Calculate the “surface pseudo liquid density” of the following reservoir

composition.

Component Mole percent

Methane 44.04

Ethane 4.32 Properties of

Propane 4.05 heptane +

Butane 2.84 API gravities = 34.2

Pentane 1.74 SG = 0.854

Hexane 2.9 Mol wt = 164

Heptane + 40.11

SOLUTION EXERCISE 2

Estimate 44.65 lb/cu ft. 0.716 gm/cc lb/cuft

From fig 12 Density 0.326 20.3424

C1

Density 0.47 29.328

C2

Component Mole Mol Weight Liq Liquid

fraction Weight Density Volume

lb/lb lb at 60 F &

mole 14.7 psia

lb/cu.ft cu ft.

z M zM o zM/ o

Methane 0.4404 16 7.0464 20.3424 0.34639

Ethane 0.0432 30.1 1.30032 29.328 0.04434

Propane 0.0405 44.1 1.78605 31.66 0.05641

Butane 0.0284 58.1 1.65004 35.78 0.04612

Pentane(n&i) 0.0174 72.2 1.25628 38.51 0.03262

Hexane(n&i) 0.029 86.2 2.4998 41.43 0.06034

Heptane+ 0.4011 164 65.7804 53.26 1.23508

Total 1 81.31929 1.8213

Density = 81.32 lb / 1.82 cu ft

= 44.65 lb/cu.ft

This trial and error method is very tedious so Standing and Katz devised a chart which

removes the trail and error required in the calculation. The densities have been con-

verted into the density of the heavier components, C3+

, and the weight percent of the

two light components, methane and ethane in the C1+

and C2+

mixtures. Figure 13.


70

60

50

40

30

10

20

30

40

50

60

70

De

nsity o

f syste

m in

clu

din

g m

eth

an

e a

nd

eth

an

e,

lb/c

u f

t

De

nsity o

f p

rop

an

e p

lus,

lb/c

u ft

Wt %

eth

ane in

eth

ane p

lus m

ate

rial

01020304050

Wt %

met

hane

in e

ntire

sys

tem

0

10

20

30

Figure 13 Pseudo-liquid density of systems containing methane and ethane 10.

We shall examine through examples various ways of calculating downhole reservoir

fluids densities dependant on the data available.

The three considered are:

1. The composition of the reservoir fluid is known.

2. The gas solubility , the gas composition and the surface oil gravity is known

3. The gas solubility, and gas and liquid gravities are known.

1. The composition of the reservoir fluid is known.

The procedure is illustrated using the following two exercises .


22

EXERCISE 3.

Calculate the surface density of the mixture in exercise 2 using the chart of figure 13

The pseudodensity is converted to reservoir conditions firstly by taking the effect

of pressure and secondly accounting for the effect of temperature. The variation of

density with respect to pressure and temperature has been investigated and it has

been demonstrated that thermal expansion is not affected by pressure. Standing &

Katz took National Petroleum Standards data and with supplementary data produced

correction factors for pressure and temperature to convert atmospheric density to

reservoir density.

The compressibility and thermal expansion effects have been expressed graphically

in Figures 14 and 15.

10

9

8

7

6

5

4

3

2

1

025 30 35 40 45 50 55 60 65

Density at 60ºF and 14.7 psia, lb/cu ft

De

nsity o

f p

ressu

re m

inu

s d

en

sity a

t 6

0ºF

1

4.7

psia

lb

/cu

ft

Pressure, psia

15,0

00

10,0

00 8

,000

5,000

6,000

4,000

3,000

2,000

1,000

Figure 14 Density correction for compressibility of liquids 8.


10

9

8

7

6

5

4

3

2

1

025 30 35 40 5045 55 60 65

Density at 60ºF and pressure P, lb/cu ft

De

nsity a

t 6

0ºF

min

us d

en

sity a

t te

mp

era

ture

, lb

/cu

ft

80

100

120

160

180

200

220

Temperature ºF

240

140

60

Figure 15 Density correction for thermal expansion of liquids 10.

EXERCISE 4.

Calculate the density of the reservoir liquid of exercise 3 at a reservoir temperature of 5,500 psia and 180 oF

Full compositional data may not always be available and the characterisation of the

produced fluids will vary from full compositional analysis to a description of the

fluids in terms of gas and oil gravity. The procedure just described is for the situa-

tion where the composition of the reservoir fluid is known. The procedures which

follow cover the situation where a less comprehensive analysis is available. These

methods make use of empirical correlations.


24

2. Reservoir Density when the Gas Solubility , the gas composition and the surface

oil gravity are known

By considering surface liquid as a single component and knowing the composition

of the collected gas the techniques previously discussed can be used to determine

reservoir liquid density. Again we will illustrate the procedure with an example

EXERCISE 5.

A reservoir at a pressure of 4,000 psia and a temperature of 200oF has a producing gas to oil ratio of 600 scf/STB. The oil produced has a gravity of 42 oAPI. Calculate the density of the reservoir liquid. The produced gas has the following composition

Component Mole Fraction Methane 0.71 Ethane 0.13 Propane 0.08 Butane 0.05 Pentane 0.02 Hextane 0.01

3. The Gas Solubility, and Gas and Liquid gravities are known.

Katz has produced a correlation (figure 16) to enable densities to be determined when

the only information on the gas is its solubility and its gravity. The figure gives ap-

parent liquid densities of gases against gravity for different API crudes

0.615

20

25

30

35

40

45

0.7 0.8 0.9 1.0 1.1 1.2 1.3 1.4

Gas Gravity

Ap

pa

ren

t L

iqu

id d

en

sity o

f D

isso

lve

d G

as a

t

60

F

an

d 1

4.7

psia

, lb

/cu

. ft.

20 API Crude

30

40

50

60

Figure 16 Apparent liquid densities of natural gases